Design and Analysis of Capacitor Buck Power Supply Circuit
description:
Capacitor buck power supply
The conventional method of converting AC mains to low-voltage DC is to use a transformer to step down and then rectify and filter. When limited by factors such as volume and cost, the simplest and most practical method is to use a capacitor buck power supply.
First, the circuit schematic:
The basic circuit of the capacitor buck simple power supply is shown in Figure 1. C1 is the step-down capacitor and D2 is the half-wave rectification diode. D1 provides the discharge circuit to C1 during the negative half cycle of the mains, D3 is the Zener diode, and R1 is the shutdown. The charge bleeder resistance of C1 after power supply. The circuit shown in Figure 2 is often used in practical applications. When it is necessary to supply a large current to the load, the bridge rectifier circuit shown in Figure 3 can be used. The unregulated DC voltage after rectification is generally higher than 30 volts, and will fluctuate greatly with the change of load current. This is because the internal resistance of such power supply is very large, so it is not suitable for high current supply. Application.
Second, the device selection:
1. When designing the circuit, first determine the exact value of the load current, then refer to the example to select the capacity of the step-down capacitor. Since the current Io supplied to the load through the step-down capacitor C1 is actually the charge and discharge current Ic flowing through C1. The larger the C1 capacity is, the smaller the capacitive reactance Xc is, and the larger the charge and discharge current flowing through C1. When the load current Io is less than the charge and discharge current of C1, the excess current will flow through the Zener. If the maximum allowable current Idmax of the Zener is less than Ic-Io, the regulator will burn out.
2. To ensure reliable operation of C1, the withstand voltage selection should be greater than twice the supply voltage.
3. The bleeder resistor R1 must be selected to vent the charge on C1 for the required time.
Third, the design example
In Figure 2, C1 is known to be 0.33μF, and the AC input is 220V/50Hz. Find the maximum current that the circuit can supply to the load. The capacitive reactance Xc of C1 in the circuit is: Xc=1 / ( 2 πf C ) = 1/ (2*3.14*50*0.33*10-6)= 9.65K The charging current (Ic) flowing through capacitor C1 is: Ic = U / Xc = 220 / 9.65 = 22mA. Generally, the relationship between the capacity C of the step-down capacitor C1 and the load current Io can be approximated as: C = 14.5 I, where the capacity unit of C is uF and the unit of Io is A.
Capacitor buck power supply is a non-isolated power supply. In application, special attention should be paid to isolation to prevent electric shock.
Pay attention to the following points when using capacitor step-down:
1 Select the appropriate capacitor based on the current of the load and the operating frequency of the AC, not the voltage and power of the power supply.
2 Current-limiting capacitors must use non-polar capacitors, and electrolytic capacitors must never be used. Moreover, the withstand voltage of the capacitor must be above 400V. The most ideal capacitor is an iron-shell oil-immersed capacitor.
3 Capacitor buck cannot be used in high power conditions because it is not safe.
4 Capacitor buck is not suitable for dynamic load conditions.
5 Similarly, capacitor buck is not suitable for capacitive and inductive loads.
6 When DC operation is required, the power supply should be half-wave rectified as much as possible. Bridge rectification is not recommended. And the power supply must meet the conditions of a constant load.
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